 Well, we were to consider what is called a Cauchy's criteria for uniform convergence of a sequence of functions. So, let us begin with that Cauchy's criteria for uniform convergence. So, as usual we will take a sequence of functions f n from e to r and we shall say that this sequence of functions satisfies this Cauchy's criteria. So, this f n is said to satisfy Cauchy's criteria if for every epsilon bigger than 0 there exists n 0 in n such that for all n and m bigger than or equal to n 0 such that n and m bigger than or equal to n 0 this implies mod f n x minus f m x is less than epsilon and this should happen for every x in d. Suppose we had simply stopped here it would have simply meant that this sequence of real numbers f n x that is a Cauchy's sequence, but what we want is something more what should happen is that for every x in e this mod f n x minus f m x should be that means this n 0 should be again independent of x that is n 0 which should work for every x such that mod f n x minus f m x is less than epsilon. Sometimes this sequence is also such a sequence is also called uniformly Cauchy we also say that f n is a uniformly Cauchy's sequence. Now, what does what does the criteria say that if you if you are given a sequence of functions then that converges uniformly if and only if it satisfies this Cauchy's criteria. So, that is the theorem let f n from e to r be a sequence of functions then f n converges uniformly if and only if f n satisfies Cauchy's criteria or in other words if and only if f n is uniformly Cauchy f n satisfies Cauchy's criteria. Now, before going to the proof of this let me again say something which I said yesterday that is we had defined yesterday this set b of e this set of all bounded functions on e and on that we have defined a norm and metric induced by the by the norm and that metric we have called d infinity. So, what is let us recall what is d suffix infinity that is d suffix infinity of f and g this is nothing but norm of f minus g that is or which is same as supremum of mod fx minus gx for x in e and that is how that is a metric and is it is it clear that saying that our sequence is uniformly Cauchy or saying that the sequence satisfies Cauchy's criteria it is same at saying that this sequence is a Cauchy's sequence in this metric. Because what this means is a suppose you take the supremum over x in e then this says that d d between f n and f m is less than epsilon distance between f n and f m is less than epsilon in that metric. So, of course you know that in any metric space if a sequence converges it is a Cauchy's sequence that is true in any metric space converges true if it is a complete metric space. So, this theorem essentially says that this is a complete metric space. Now, let us go to the proof this part is relatively easy to prove that is if f n converges to f then it satisfies Cauchy's criteria again as I said to show that if a sequence converges it is a Cauchy's sequence that is a straight forward straight forwarding. But anyway let us let us go through it suppose f n converges suppose f n converges converges to some f uniformly then we have to show that it is a Cauchy's sequence to show that it is a Cauchy's sequence take some epsilon and show that such an n 0 exists. So, let epsilon be bigger than 0 then since we know that f n converges to f uniformly we can use that there exists n 0 in n such that for all n bigger than or equal to n 0 we have mod of f n x minus f x less than I can say less than epsilon by 2 and this should happen for every x in e for every x and this is a standard way of going now you take any n n m bigger than or equal to n 0 and look at f n x minus f m x. So, now consider mod of then mod of f n x minus f m x what is to be done is clear added subtract f x. So, this is less than or equal to mod of f n x minus f x plus f x minus f m x and each of this quantity is less than epsilon by 2. So, this is less than epsilon by 2 plus epsilon by 2 and that is equal to epsilon and again since this n 0 was independent of x this works for every x in e. So, whatever we have done about that n 0 should work for every x in e. So, this also is true for every x in e so this shows that f n satisfies Cauchy's criterion that is what is meant by what we want to show is that for every epsilon there exists n 0 such that whenever n and m are bigger than or equal to n 0 mod of f n x minus f m x is less than epsilon for every x in e. So, that is what we have shown now next is the other way now we assume that f n satisfies Cauchy's criterion and from that we need to show that it is a uniformly convergent sequence. So, suppose f n satisfies Cauchy's criterion now first of all what we can do is that take any x in e and consider this sequence f n x consider this sequence f n x then that is a Cauchy's sequence. So, for each x in e this is satisfied so it is same as say that this is a Cauchy's sequence of real numbers. So, then f n x is a Cauchy's sequence of real numbers. So, real numbers and real line is complete. So, once a sequence is Cauchy's sequence of real numbers it should converge to some real number that real number we shall call f x. So, and hence converges some real number we call that real number f x. So, this is this can be done for each x in e this can be done for each x in e. So, that means this defines a function from e to r. So, that is that is this defines a function f from e to r and what is the how the function is related to the sequence f n each f x is a limit of this f n x each f x is a. So, it is given by this f of f of x is equal to limit n that is to infinity f n x. But at this point we have only said that it is point wise converges because how did we construct f we started with some x and said that the corresponding sequence is a Cauchy's sequence of real numbers and hence that has a limit that limit we have taken and that limit we have called f x. So, with this construction we have only got a function f which is a point wise limit of this sequence f n. But that is not what we want what we want is a limit should be uniform limit. So, obviously if at all uniform limit exists that has to coincide with point wise limit it cannot be some different function. So, we should try to show that this is a uniform limit. So, that means we get but this is we can write this as a claim what is the claim we want to show that f n converges to f uniformly f m converges to f uniformly. Now, let us take again some epsilon bigger than 0 then what we can say is that since we already know that f n is a uniformly Cauchy's sequence that is f n satisfies Cauchy's criteria using that there exists some n 0. So, there exists some n 0 in n such that m and n bigger than or equal to n 0 this implies mod of f n x minus f m x less than epsilon and again this should happen for every x in e that is the same n 0 should work for every x in e. Now, the argument which is given after this is a fairly standard one what we say after this is that this is true for every m and n bigger than or equal to n 0. So, what we do after this is keep n fixed or one of them either m let us say suppose keep n fixed and let m vary. Let m vary means consider all possible values of m bigger than or equal to n 0 that means let m go to infinity if m goes to infinity what should happen this should go to f x this should go to f x, but this inequality is true for every m that is mod f n x minus f m x is less than epsilon is true for every m. So, in the limit at the most what could happen is that this may become less than or equal to. So, this is what we can say after this is keep n fixed and let m vary from tends to infinity then since it is because of this we know that if m tends to infinity f m x goes to f x f m x goes to f x. So, this will go to f x and hence this will become then mod f n x minus f x less than or equal to epsilon, but you can say this less than or equal to does not matter here itself we could have taken some numbers smaller than epsilon and then we could have got less than epsilon. So, that is the minor thing so, this means that f n converges to f uniformly that is given epsilon we have found some n 0 such that whenever n is bigger than that n 0 mod f n x minus f x is less than epsilon of course, this again happens for every x in e. So, again let me again comment that what this theorem actually means is that that space B e which we have taken is a complete metric space. Now, translating this to the convergence of a series we get one useful criteria for convergence of a series it is it is called Weierstrass m test Weierstrass is a name so Weierstrass m test this is a test which helps in deciding easily about certain kinds of series which are uniformly convergent or not. So, again what we do is let us let us first see what the test says the test is as follows again we take a sequence f n f n from e to r and suppose we have some numbers m n let us say such that mod of f n x is less than or equal to m n for all x in e. So, suppose in our language this is same as saying that norm of f n is less than or equal to m n because mod of f n x less than or equal to m n for each x in e this is essentially same as saying that if you look at this norm of f n suffix infinity this is less than or equal to m n. So, in particular you can take this m n as this number itself you can take m n as this number itself and this is something that you can do whatever this is the sequence given once f n is bounded we can always talk of this. So, we can always take this m n, but what we what is something more and now m n is a sequence of real numbers m n is a sequence of real numbers. So, suppose this happens and if you look at the series sigma m n n going from 1 to infinity converges then the functional series sigma f n that is series of functions then sigma f n n going from 1 to infinity converges uniformly that is what again recall what is involved there should exist some number m n. In fact this will be always a non negative real number such that for every x in e mod f n x is less than or equal to m n and the series sigma m n should be convergent that is the series of positive numbers and that is fairly easy to check because there are several tests for checking the convergence of the series of positive numbers. So, that can be checked very easily and then if that happens we can conclude from this vice truss m test that sigma f n converges uniformly. For example, suppose something like this you take the series sigma n going from 1 to infinity let us say sin n x by n square that is a series of functions and what is the obvious choice of m n here 1 by n square because each of this is not sin n x is less than or equal to 1. So, each of this is less than or equal to 1 by n square and sigma 1 by n square converges that is a well known thing sigma 1 by n square converges. So, this series of functions converges uniformly and this way you can check the convergence uniform convergence of a series in a fairly easy manner. So, that is the idea of vice truss test these numbers m n you can calculate in most of the cases. Again remember we do not have to have m n to be exactly equal to this it can be any number bigger than or equal to this all that we need subsequently is that sigma m n should be convergent. So, let us look at the proof of this test it depends basically on what we have proved just now that a sequence converges if and only if it satisfies the Cauchy criteria. So, all right there are as you know anything that you want to talk about a series has to be talked in terms of its partial sums. So, let us say let us take S n is equal to f 1 plus f 2 plus f n and similarly let us look at the partial sum of this let us look at the partial sum of this suppose I call that as t n this is these are numbers m 1 plus f 2 etcetera plus m n. Now, to show that this series converges uniformly it is same as saying that this sequence S n converges uniformly that is what we want to show right. So, this is what we want to show S n converges uniformly. Now, by using the last theorem we have seen that to show that the sequence converges uniformly it is sufficient to show that it satisfies the Cauchy's it satisfies Cauchy's criteria. So, that is what we shall show we shall show that it satisfies Cauchy's criteria all right. Now, to look at the Cauchy's criteria we will have to look at S n minus S n right that is we have to show that for every epsilon there exist n 0's that whenever n and m are bigger than or equal to n 0 mod of S n x minus S n x is less than epsilon. So, first of all let us calculate what is S n x and S n x are. So, S n x minus S n x this is same as S n x is because S n x is same as f 1 plus f 2 plus f n and S n x is same as. So, f 1 plus f 2 plus f n. So, let us say n bigger than m suppose we take n bigger than m then this will be same as f m plus 1 plus f m plus 2 etcetera to f m plus n. So, we can say that this is same as sigma let us say j going from m plus 1 to n f j x. So, let us take the absolute value then this will be less than or equal to mod f j x right and what do you know about this mod f j x that mod f j x is less than or equal to m j for every x in each. So, I can say that this is less than or equal to sigma j going from m plus 1 to n m j for every x in each. Now, what can you say about this number sigma j going from m plus 1 to m n we know that sigma m n is convergent sigma m n is convergent that is same as saying that T n is. In fact, this is nothing but this number is nothing but T n minus T n right and since sigma m n is convergent is same as saying that T n is convergent right, but if T n is convergent it is Cauchy it is a sequence of real numbers. So, now the argument after this is standard what you can say is that given any epsilon you can always find n 0 such that whenever n is bigger than or equal to n and m are bigger than or equal to n 0 T n minus T m is less than epsilon and if what T n minus T m is less than epsilon it shows that mod s n x minus s m x is less than epsilon for every x in each which is same as saying that this essence satisfies Cauchy's criteria and hence converges uniformly. So, so far we have just discussed what is meant by uniform convergence of sequence or series and how to check in particular Cauchy's criteria and this Weierstrass m test. By the way this Weierstrass m test is a very practical and useful test and using this you can check the uniform convergence of the series of functions in several cases. Now, let us go back to our earlier problem what we wanted to know was that if each f n has certain properties then does it follow that the limit f also has the same properties at least under the condition of uniform convergence. We have seen that under the point wise convergence that is not true. So, let us now check about uniform convergence. Now, if you look at the properties like continuity then see till now we have taken this set e to be any arbitrary set we did not bother about what that set e was, but to consider the properties like this we if we want to talk about the limit or continuity etcetera then the whole thing has to happen inside some metric space. So, now we will take the set e to be a subset of a metric space. So, so let us say that let us say that x d is some metric space and of course, subsequently when we want to discuss the differentiability integrity etcetera then we will have to take it as a subset of a real line in fact an interval, but for the continuity still we can take any arbitrary set in a metric space. So, suppose e is a subset of x and f n is from e to r and f is also from e to r and f is also r and suppose f n converges to f uniformly then what we want to say is that if each f n is continuous then f is also continuous, but instead of proving that directly since the definition of continuity involves the limits we shall prove something about the limits and then whatever we want to say about the continuity will follow as a crawling. So, let us say that suppose this f n has a limit as some point. Now, to talk about the limit suppose I take this point x and I want to talk about the limit as f n tends to let us say as t tends to x then for that this x does not have to belong to e you know that for talking about a limit the point at which you are talking about a limit does not have to be in the domain, but what it must be is that it must be at least a limit point because in every neighborhood the function should be defined though it need not be defined at the point x etcetera. So, let us say that suppose f n tends to f uniformly and x is a limit point of e and x is a limit point of e and let limit suppose for each f n limit of f n t as t tends to x exist. Suppose for each f n limit of f n t as t tends to x exist and suppose we call that limit as l n suffix n then what we want to say is the following that this sequence l n converges and suppose l is that limit then limit of f t as t tends to x also exist and that limit is same as the limit of this l n. So, then limit l n as n tends to infinity exist that is this is a convergent sequence also limit of f t as t tends to x exist and both the limits are the same and limit of l n as n tends to infinity is same as limit of f t as t tends to x that is limit of l n as n tends to infinity that is this is n as n tends to infinity is same as limit of f t as t tends to x. Suppose we write this again l n and f t in their full form because l n is nothing but limit as t tends to x of f n t. So, suppose I write that here that is limit the left hand is same as limit as n tends to infinity of l n is limit as t tends to x of f n t and this is equal to limit as t tends to x and f t is nothing but limit of f n t as n tends to infinity because that is what we assumed that f n converges to f. So, f t is nothing but limit as n tends to infinity of f n t. So, if you look at this last equation you will understand what I said right in the beginning by discussing about uniform convergence what we are saying is that if f n converges to f uniformly these limits can be interchanged here you are taking limit as t goes to x first and then limit as n tends to infinity whereas it is here it is other way here you are taking limit as n tends to infinity first and then taking the limit as t goes to x of the same quantity f n t. Now, let us prove these things one by one first thing to prove is that l n converges that is limit as n goes to infinity l n exists. But l n is a sequence of real numbers l n is a sequence of real numbers. So, to show that it converges it is sufficient to show that it is quotient and that we can show by using that since f n converges to f uniformly it is also a uniformly Cauchy sequence. So, let us let us start with that. So, let us say let epsilon be bigger than 0 then there exists n 0 in n such that n and m bigger than or equal to n 0 this implies mod f n t minus f m t this is less than epsilon this less than epsilon for every t in e for every t in e. Now, since this is true for every t in e we can let t goes to x this inequality satisfied for every t. So, it will be and we know that limit of this as t goes to x exist that is l n limit of this as t goes to x exist and that is l m. So, l n minus l m may not be strictly less than or equal to epsilon in the limit at the most the inequality need not remain strict, but it will remain less than or equal to. So, we can say that letting t tends to x letting t tend to x we have mod l n minus l m is less than or equal to epsilon. Now, that shows that this is a Cauchy sequence that shows this is a Cauchy sequence. So, this implies that l n is a Cauchy sequence l n is a Cauchy sequence and since it is a Cauchy sequence of real numbers it is a Cauchy sequence of real numbers and hence it should converge to some real number and hence converges to some say real number l belonging to R. So, we have shown this first requirement that limit of l n exist and suppose now and we have called it limit as l. Now, what remains to show that that number l is same as this limit of f t as t tends to x let us go to that now. So, to show l is equal to or let me write this limit as limit of f t as t tends to x is equal to l by the way it is a standard convention in analysis you would have come across that is this equation means that this limit exist and is equal to l it means both the things that this limit exist and limit is equal to l that is the meaning of this symbol. So, once we show this it is that once we show this then everything is proved that is once we show this this equation is proved this conclusion is proved and this is nothing but rewritten we have rewritten the same thing in a different manner. So, this is nothing new now the proof of this equation is again follows a very standard technique used quite frequently in analysis sometimes known as 3 epsilon proof or epsilon by 3 proofs etcetera. Basic ideas you just add and subtract some terms see what is the idea we want to show that the limit of l limit of f t as t tends to x is equal to l. Now, by definition what does it mean that given epsilon you should be able to find some delta such that whenever distance between t and x is less than delta distance between f t and l is less than epsilon. Now, that means that means what does it mean practically we want to show that this difference f t minus l is small this difference can be made arbitrarily small that is what we want to show by taking t as close to x as you want. Now, we know nothing about the difference mod f t minus l, but we know the difference about several other things by whatever we assume. So, basically we add and subtract those terms for example, we know that mod f t minus f n t can be made small. For example, I can write this we know that mod f t minus f n t this can be made small because f n converges to f uniformly then we also know that f n t minus l n that can be made small because limit of f n t as t goes to x is l n. So, if you go t to x this can be made small and then finally, if you look at l n minus l this also can be made small if l n is close to if l n is sufficiently large the difference between l n and l is small. So, what we can say now is this for any t for any t this much is true that mod f t minus l is less than or equal to mod f t minus f n t plus mod f n t minus l n plus mod l n minus l. So, we want to choose delta in such a way that this becomes less than epsilon. So, we shall make the arguments in such a way that each of this term becomes less than epsilon by 3 or if you make each of is less than epsilon the whole thing will become less than 3 epsilon that also does not matter. So, that is why such a thing is called a 3 epsilon proof or epsilon by 3 proof or whatever basic idea here is that you add and subtract some terms. Now, let us look at 1 by 1 when will this be small mod f t minus f n t this will be. So, we will begin with this let epsilon bigger than 0 be given we want to make this whole some less than epsilon. So, we will try to make this whole thing this less than epsilon by 3 this less than epsilon by 3 and this is less than epsilon by 3. Now, out of this this and this will become less than epsilon by 3 if n is sufficiently big. So, that choice we will make first that choice we will make first. Now, we can say that there exists let me call to begin with there exists n what if n such that n bigger than or equal to n 1 implies mod f t minus f n t is less than epsilon by 3 is less than epsilon by 3, but one more thing is important this happens because f n tends to f uniformly this happens for every t in e this happens for every t in e. Then since l n converges to l we also say that there exists n 2 in n such that n bigger than or equal to n 2 implies mod l n minus l is less than epsilon by 3. So, now, if you take some n which is bigger than or equal to both of these say maximum of n 1 and n 2 then for that n both these inequalities will be satisfied. So, consider some such n. So, let so consider some n bigger than or equal to maximum n 1 and n 2 then for that n this is less than epsilon by 3 for every t in e and this is also less than epsilon by 3. What remains is this what is to be done about this now we will now look at only for that particular n this n which we have chosen this n which we have chosen for that n look at f n t minus l n. Now, f n t minus l n will become small if we if t is sufficiently close to x if because that is what we have showed that limit of f n t as t goes to x is. So, now, what you do is choose delta in such a way that whenever distance between t and x is less than delta this difference is less than epsilon by 3. So, we can say now what we can say is that for this n since limit of f n t as t goes to x is equal to l n there exists delta bigger than 0 such that wherever the distance between t and x is less than delta distance between t and x is less than delta this implies mod f n t minus l n is less than epsilon by 3. Now, the proof is more or less over what so given epsilon we have found a delta such that whenever distance between t and x is less than delta mod f n f n t minus l n is less than epsilon by 3 mod f n t minus l n is what is this n what is this which n or what is this f n this n is some number which is bigger than this n 1 and n 2 this n 1 and n 2 we have already chosen. Now, once this is the case since for this n is bigger than or equal to n 1 and n 2 for that particular f n mod f t minus f n t is less than epsilon by 3 this is less than epsilon by 3 and this is also less than epsilon by 3. So, this whole thing is less than epsilon. So, that means we have given epsilon we have found a delta such that whenever distance between t and x is less than delta mod f t minus l is less than epsilon that is same as saying that that is same as showing this. So, that proves this theorem completely again what is the theorem that if the if the each of f n has a limit as t goes to x then f also has a limit as t goes to x and that limit is nothing but limit of this l n. Now, as a simple corollary we can say that if each f n is continuous then f is also continuous. So, we can say that as a corollary of this under the same hypothesis that what is hypothesis that suppose f n tends to f uniformly suppose f n tends to f uniformly then in fact this can be stated even at a point, but let us say what a point if f is continuous if f if each f n is continuous f n is continuous either we can say at some point x in e or at all points x in e it makes no difference in the proof. Let us say if each f n is continuous to begin with let us talk of some point continuous at some x in e, but this time I am taking x in z there I had taken just the limit point x in e then f is continuous at x in e and in particular if this happens for all points that is that means if f n is continuous at all points in e f is also continuous at all points in e. So, in particular if do you use this notation c of e you are following set of all continuous real valued functions on e that is c of e that is f from e to r f is continuous. So, instead of saying inverse that each f n is continuous at all points x in e I will simply say if f n belongs to e for all n f n belongs to e for all n then f belongs to not not f n e c of f n belongs to c of e for all n then f belongs to c of e. In fact, this says something about this space c of e, but I will come to that little later first of all let us prove this scrawling you can say that proving this scrawling is nothing really different that is suppose let us let me again look at this instead of say now here instead of say that x is the limit point of e we are now taking x belongs to e we are now taking x belongs to e. Now, what is the meaning of saying that f n is continuous at x it means that this limit of f n t is t tends to x that is nothing but f n of x then. So, this l n this will be changed to f n of x and what is the conclusion it means that that the limit of f n x exist then limit of f t as t tends to x exist and that limit that limit is nothing but see remember this l n is nothing but now l n is nothing but f n of x limit n tends to infinity that is limit as t tends to x of f t that is same as limit n tends to infinity f n of x. But, what is limit n tends to infinity f n of x that is same as f x because we have assumed that f n converges to f. So, this is nothing but f of x so that means limit as t tends to x of f t is same as f of x which is same as saying that f is continuous at x. So, essentially all that we do is we apply this theorem at the point x we apply this theorem at the point x and then get that if each f n is continuous f also must be continuous only I use this if f n is continuous at x is same as saying that this l n is nothing but f n of x and you just use that here you get that limit of f t as t tends to x is f x that shows that f is continuous at x. Further if this happens for all points x in e then f also must be continuous at every x in e that is the things say nothing extra that is once we show that f is continuous at if each f n is continuous at some x in e then f is also continuous at x in e. So, if the first statement is true for every x in e this also should be true for every x in e that is all now coming back to this notation let me again go back to this metric space set of all functions which are bounded in e with respect to this matrix d suffix infinity we have shown that this is a complete metric space that was the beginning. Now in general if you take any arbitrary set e in a metric space and then every continuous function need not be bounded, but if you put an extra condition that e is a compact set then you know that every continuous function is bounded. So, now let us consider that suppose e is compact then this c of e then saying that every continuous function is bounded it is same as saying that c of e is a subset of b of e then c of e is a subset of b of e do you agree. In fact this is nothing but saying that every continuous function on compact set is bounded this is nothing but saying in the words since e is compact every continuous real value function is bounded. Now what does this theorem say about c of e look at this last statement if each we have already seen that uniform convergence is nothing but convergence in this metric uniform convergence is nothing but a convergence in this metric. So, what does this theorem say that suppose you take a sequence f n in c e and if f n then obviously each of this f n must be in b e and if f n converges to f in this metric then that f also belongs to c of e what does that mean that this c of e is a close subset that c of e is a close subset of this metric space. But suppose you know that metric space is complete and if a subset is closed then what can you say about that subset that is also complete. Now you cannot say it is compact but it is it is complete you can always say it is complete. So, a close subset of a complete metric space is again complete. So, what this theorem says or what this chronology says that c of e is a close subspace or close subset of b of e. In fact, we can also say close subspace because the c of e if you regard this b of e as a vector space c of e is also a subspace because c of e is also a vector space it is a subspace and it is a close subset or if you want you can say subspace of b of e and hence complete. That means what we have shown is that means if you look at this space whenever e is compact if you look at this space c of e with respect to this metric this is a complete metric space this is a complete metric space in particular spaces like c of a b with this metric supreme metric given by the supreme now this is a complete metric space. Now sometimes we are also interested in knowing see we have seen an example that every uniformly convergent sequence is point wise convergent, but the converse is false in general the converse is false. So, sometimes we also want to know what additional conditions can be put on a point wise convergent sequence. So, as to make it uniformly convergent there is one very well known theorem in that direction that is called denies theorem. What it says is that if the sequence is monotonically increasing or monotonically decreasing and if it is point wise convergent then it is also uniformly convergent, but this is something we shall prove in the next class.